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I am trying to design a single transistor RF power amplifier. For a single transistor amplifier, the voltage at the base/gate of the transistor determines the current through the transistor and ultimately the output power. Then why is power input to the transistor maximized by impedance matching. Why can't impedance bridging be done to increase the voltage at the gate/base of the transistor and hence increase the collector/drain current?

Edit: If your answer is maximum power transfer, I dont understand how impedance matching ensures maximum power transfer. I understand that impedance matching ensures that maximum power enters the transistor but the transistor is not a power amplifier. It is usually a transconductor or a voltage amplifier. How does the output power maximize when the input power is maximized?

Edit again: I realized that this question Why do we care about matching the input impedance of receiving RF amps? is very similar to the one I am asking. The accepted answer there is that maximizing the power input to the amplifier through impedance matching increases the SNR of the input signal thereby increasing the SNR of the output signal too. Is the same argument applicable here too? Is maintaining the SNR of the signal the sole reason to maximize the input power to the amplifier by impedance matching?

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    \$\begingroup\$ You mean increasing the load impedance seen by the source driving the input of the amplifier ? Instead of matching ? \$\endgroup\$ – Meenie Leis Mar 18 '18 at 18:27
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    \$\begingroup\$ Impedance matching is there to maximize power transfer. That isn't always a goal, though. Amplifying the signal from a microphone, for example, is more about operating the microphone for maximum fidelity of the audio that is retrieved from it and not about getting the most power transferred. But if you are a ham radio operator and have an existing transmitter and you want to add a linear amplifier using a 4CX1000A, then you want maximum power transfer and therefore impedance matching. \$\endgroup\$ – jonk Mar 18 '18 at 19:09
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    \$\begingroup\$ The input needs to be 50 Ohms to prevent a high VSWR. At RF you can't just have a high impedance input. Reflections and such. \$\endgroup\$ – mkeith Mar 19 '18 at 4:22
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    \$\begingroup\$ You have to match the impedance in order to maintain signal integrity. If you are using a totally unmodulated sine wave, then I guess signal integrity is not such a big deal. It will remain an unmodulated sine wave no matter what abuse you put it through. But if it is modulated, then you cannot have reflections because then echos of past signals will interfere with present signals. This will destroy the integrity of the baseband signal. So you do not have broad latitude to ignore impedance matching. \$\endgroup\$ – mkeith Mar 20 '18 at 3:58
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    \$\begingroup\$ I didn't fully understand that answer. But it sounds good! Reflections from old signals are also a form of noise, so they degrade SNR. It is not random noise, but it can still be considered noise in the sense that it is not part of the intended signal. \$\endgroup\$ – mkeith Mar 20 '18 at 4:18
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Impedance matching is important in RF circuits to prevent reflections. When the input impedance of the load matches the characteristic impedance of the transmission line there is no reflection at the load.

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  • \$\begingroup\$ What harm do such reflections do? Do they corrupt the input signal? Do they damage the input source? Or do they just cause loss of power transferred? I am not sure about the reason reflections are a worry in RF circuits \$\endgroup\$ – Kevin Selva Prasanna Mar 20 '18 at 3:21
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    \$\begingroup\$ @KevinSelvaPrasanna They distort the signal. Imagine you have a single pulse signal. It reflects at the open-circuit load, travels back to the source, reflects there, travels back to the load, etc. So the load sees many pulses, even if only one was sent. \$\endgroup\$ – τεκ Mar 20 '18 at 3:34
  • \$\begingroup\$ Wow, cool. Thanks. I found another answer at electronics.stackexchange.com/questions/288645/… which talks about SNR of the signal. Is that applicable here too or is reflections the major concern? \$\endgroup\$ – Kevin Selva Prasanna Mar 20 '18 at 3:58
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Impedance bridging is done if you are interested in Voltage rather than Power. It will ensure that maximum voltage is driven to the amplifier input with minimum signal degradation. It consumes less input power, as the current drawn by the series combination of source and load impedances is low. It is relevant in voltage amplifiers.

Impedance matching is done when you are interested in Power. It is relevant in Power amplifiers, where output loads are of low resistances. Here the input impedance is equal to the source impedance. The amplifer will now extract maximum Power off the source at its input (MPT-Theorem). Compared to previous case, the input current of the amplifier is higher too. Hence, the amplifier can now drive higher current in the output collector circuit. i.e., More Power can be extracted at the output. Also, it is not just about the output Power. It is also about Power efficiency in terms of input power and output power of the amplifier. Maximum Power efficiency is obtained only when the impedances are matched.

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  • \$\begingroup\$ A single transistor amplifier is essentially a voltage amplifier. The collector current depends on the base voltage alone assuming the emitter is grounded. Why is impedance matching recommended in single transistor power amplifiers if it is the output characteristics are a function of the input voltage alone and not the input power? \$\endgroup\$ – Kevin Selva Prasanna Mar 20 '18 at 2:48
  • \$\begingroup\$ And why do you talk about efficiency in terms of signal power? Efficiency is usually discussed in term of the power supply. Why do we care about the signal power wastage which is a tiny fraction of the power delivered by the power supply? \$\endgroup\$ – Kevin Selva Prasanna Mar 20 '18 at 3:17
  • \$\begingroup\$ In voltage amplifiers , nobody says that impedance matching is necessary. Its relevant only relevant in power amplifiers. \$\endgroup\$ – Meenie Leis Mar 20 '18 at 7:25
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    \$\begingroup\$ No output current will increase. Because eventhough input voltage is lower compared to previous case, input current (Base current) has increased as the input impedance of the amp is now lower too (V/Z = I ). And its in series with source impedance (which is still the same as previous case). This is actually proof of MPT theorem how both V and I are balanced by load impedance to maximize the power . \$\endgroup\$ – Meenie Leis Mar 20 '18 at 12:01
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    \$\begingroup\$ polytechnichub.com/difference-voltage-amplifier-power-amplifier -- higher collector current due to impedance matching -- higher power at output. \$\endgroup\$ – Meenie Leis Mar 20 '18 at 12:05
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View the transistor as a box that provides transconductance between input/base and output/collector. For maximum output at the collector, the base needs the maximum input voltage.

In the presence of various reactances at the base, such as Cemitter_base and Ccollector_base (which will vary as collector resonances change with temperature and the collector VPP changes), the designer still needs to achieve some guaranteed VinPP on the base.

You may need to de-Q (include some dampening) if your RF signal must cover a wide range of frequencies.

You need to examine various stability factors, and place the various resonances at frequencies, and with adequate dampening, that you the designer can guarantee the RF amplifier will not oscillate nor will become peaky enough to cause unacceptable variations in output VPP.

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  • \$\begingroup\$ I understand your answer but I still don't understand the need for impedance matching. According to your first paragraph, to maximize output, the input voltage has to maximized. Then why is impedance matching done which does not ensure maximum voltage at the input. I understand that the parasitic reactances in the circuit influence the input and output voltage but I still fail to understand the case for impedance matching. \$\endgroup\$ – Kevin Selva Prasanna Mar 19 '18 at 4:26
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If you look at the base or gate of a transistor, it is both resistive and capacitive. Hence it is called input impedance. You want the input power to be put in the resistive part of the input so you have to eliminate the capacitive part. This is called impedance matching. Assume the input comes from a 50 Ω source, you have to match it to the complex impedance of the transistor. You can use a L matching network.

The same applies for the output of a transistor.

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